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Air in a piston/cylinder goes through a Carnot cycle. The high and low temperatures are 600 K and 300 K, respectively. The heat added at the high temperature is 250 kJ/kg, and the lowest pressure in the cycle is 75 kPa. Find the specific volume and pressure after heat rejection and the net work per unit mass.

Respuesta :

alimir

Answer:

Volume = 0.2688 [tex]m^{3}[/tex]/kg

Pressure = 320kPa

Net work/unit mass = 125kJ/kg

Explanation:

qH=250 (heat at high temp)

TH= 600K (high temp)

TL= 300K (low temp)

Pl= 75kPa (lowest pressure)

Efficiency = η = 1 - [tex]\frac{TL}{TH}[/tex] = 0.5 (Plugging the values from above)

So, net work done per unit mass = η*qH = 0.5*250 = 125kJ/kg

qL is the heat rejected which would be = qH - net work done = 250 - 125 = 125kJ/kg

VL (volume low) = [tex]\frac{RTL}{Pl}[/tex] = 0.287*300/75 = 1.148kJ/kg

Specific volume = VL[tex]*e^{\frac{-qL}{R*TL} }[/tex]

                           = 1.148[tex]*e^{\frac{-125*300}{.287} }[/tex] = .02688[tex]m^{3}/kg[/tex]

PH = R*TH/VH = 0.287*300/0.2688 = 320kPa

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